Question

The horizontal range of an oblique projectile is equal to the distance through which a projectile has to fall freely from rest to acquire a velocity equal to the velocity of projection in magnitude. The angle of projection is

• A. $75^\circ $
• B. $60^\circ $
• C. $45^\circ $
• D. $30^\circ $

Answer 1

Answer: (A)

Using $v^{2}-u^{2}=2aS$ , we get
$S=\frac{v^{2}}{2 g}$
Now, $\frac{v^{2} sin 2 \theta }{g}=\frac{v^{2}}{2 g}$ or $sin 2 \theta = \frac{1}{2}$
Or $sin 2\theta =sin⁡30^\circ $ or $\theta =15^\circ $
The other possible angle of projection is $\left(90 ^\circ - 15 ^\circ \right)$ , i.e., $75^\circ $ .